I had a regular combinatorcics exercise to solve and I thought it's possible to solve it in two ways but it turned out that only one way is correct. It is: A team of 4 students is to be selected for a competition. There are 8 boys and 12 girls to choose from. If the team have to include at least one boy and at least one girl, in how many ways can the team be selected?

So the simplest solution which is correct is to find the total number of possible selections (without any conditions) and then subtract from it the number of possible solutions including only boys and only girls.

But I thought the method mentioned below would be correct, it turned out to be wrong: (8 nCr 1)(12 nCr 1)(18 nCr 2) = 14688 The two first pair of parantheses mean that we need at least one boy and at least one girl of our group of 20. And (18 nCr 2) means that we need two more people to complete our team and it doesn't matter if it's a boy or a girl. Unfortuantelly, this method isn't somewhere incorrect.

The correct answer is 4280 so now I think there may be some extraneous selections in my method but I can't think of any. Would you please tell me where my approach falls?


You have chosen one boy from eight, and one girl from twelve, and then two more from the remaining eighteen children. This over-counts: for example if there are two particular boys and two particular girls then that set will be counted four times in your method.

The simple solution you give is ${20 \choose 4}-{8 \choose 4}-{12 \choose 4}$.

Another method would be ${8 \choose 3}{12 \choose 1}+ {8 \choose 2}{12 \choose 2} +{8 \choose 1}{12 \choose 3}$, giving the same correct result.

Your erroneous method is $3{8 \choose 3}{12 \choose 1}+ 4{8 \choose 2}{12 \choose 2} +3{8 \choose 1}{12 \choose 3}$, which illustrates the overcounting.

  • $\begingroup$ I think I get it now but I want to make sure. Let's use my method as an example - when I've got a set of 2 boys and 2 girls ABCD, where A and C are boys and B and D are girls and the first letter in ABCD stand for the result of (8 nCr 1), the second stand for result of (12 nCr 1) and the last two ones are from (18 nCr 2), my method would produce ABCD, CDAB, ADCB and CBAD, right? So there are unwanted permutations, because it doesn't matter in which order are they. $\endgroup$ – Vincent Dec 14 '14 at 15:38
  • $\begingroup$ And similarly if I had a team of 3 boys and 1 girl named ABCD where ABC are boys and D is a girl, my method would produce ADBC, BDAC and CDAB, so there are three times as much as I need. Did I get it right, both of my comments? $\endgroup$ – Vincent Dec 14 '14 at 15:39
  • $\begingroup$ Yes - those look correct to me. Hence the $3$s and $4$s multiplying in my last line $\endgroup$ – Henry Dec 14 '14 at 19:42

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